# If every eigenvector of $T$ is also an eigenvector of $T^{*}$ then $T$ is a normal operator

Let $V$ be a finite-dimensional inner product space over $\mathbb{C}$ and $T: V \to V$ a linear transformation.
Show that if every eigenvector of $T$ is also an eigenvector of $T^{*}$ then $T$ is a normal.

We need to show that $\forall v \in V,\ TT^*v=T^*Tv$.

I started by picking $v$ to be an eigenvector of $T$, hence $Tv=\lambda_1 v$, and $T^*v=\lambda_2 v$. Therefore: $$TT^*v=T(\lambda_2 v)=\lambda_2 Tv=\lambda_1 \lambda_2 v=...=T^*Tv$$

But I'm stuck when $v$ is not an eigenvector.

(Also, this question is taken from a chapter about Jordan forms, but I can't see how it relates)

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Have you seen that operators on $V$ can be unitarily triangularized? Have you seen that adjoint corresponds to the conjugate transpose when $T$ is represented as a matrix with respect to an orthonormal basis? – Jonas Meyer Jun 17 '11 at 9:57
@Jonas: Yes to the last question. Never heard the term 'unitarily triangularized' though. – daniel.jackson Jun 17 '11 at 10:10

If you consider matrices with respect to an orthonormal basis, what I meant in my comment about unitary triangularization is that for every $n$-by-$n$ complex matrix $A$ there is a unitary matrix $U$ such that $UAU^{-1}$ is upper triangular. An algorithm to find $U$ appears in Hogben's Handbook of linear algebra. If we forget about matrices, what this says is that given an operator $T$ on $V$, there exists an orthonormal basis $(e_1,\ldots,e_n)$ and constants $(a_{ij})_{i\leq j}$ such that $\displaystyle{Te_j=\sum_{i=1}^j a_{ij}e_i}$. In particular, $e_1$ is an eigenvector for $T$. Therefore it is also one for $T^*$, and this forces $a_{1j}=0$ for $j>1$. Then $a_{12}=0$ implies that $e_2$ is an eigenvector for $T$, thus also one for $T^*$, thus $a_{2j}=0$ for $j>2$. Since $a_{13}=a_{23}=0$, $e_3$ is an eigenvector for $T$. And so on (omitting the induction proof). Eventually it comes out that $T$ and $T^*$ are simultaneously diagonalized, so they commute.

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That's a very nice proof, thanks. Out of curiosity, can you think of one that uses Jordan forms? I'm trying to think how this relates to the material I just read. – daniel.jackson Jun 17 '11 at 10:50
I can't think of one off the top of my head that uses Jordan forms. One problem is that the Jordan form in general doesn't interact well with the inner product, so it wouldn't be easy to get information on $T^*$ from the Jordan form of $T$. It turns out that the triangular form for $T$ above is its Jordan form, but this is incidental. – Jonas Meyer Jun 17 '11 at 10:55
Ok, thanks for trying. So according to your proof, $T$ and $T^*$ can be represented by these diagonal matrices: $\text{diag}(a_{11},\ldots,a_{nn}),\ \text{diag}(\overline{a_{11}},\ldots,\overline{a_{nn}})$, and since diagonal matrices commute, so does the transformations? – daniel.jackson Jun 17 '11 at 11:07
@daniel.jackson: Yes. You're welcome. – Jonas Meyer Jun 17 '11 at 11:09

Hint for a coordinate-free proof: Begin by proving the following

${\bf Lemma.}$ If $U$ is an invariant subspace of $T\$ then $U^\perp$ is an invariant subspace of $T^*$,

then use induction on the dimension.

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Proof for lemma: let $u \in U^\perp$, then $<u, v>=0,\ \forall v \in U$. Since U is invariant, $Tv \in U$ as well, hence $<u, Tv>=<T^*u, v>=0$, therefore $T^*u \in U^\perp$. But how does this help? – daniel.jackson Jun 18 '11 at 13:38